Forgotten how to do basic high school algebra

Hi, just brushing up. Please show how you got your answers, as the answers don't mean much to me. Thanks!

1. 5n = 25

2. -4n^2 + (3n)^2

3. n^3 - 2n^2 (n + 1)

4. x3 + x2 - 9x = 9

Once again thanks for your time!

1. 5n = 25

Divide both sides by the coefficient of n.
n = 25/5 = 5

2. -4n^2 + (3n)^2

First, distribute the power of 3n, it will become 3^2 n^2. Then add the terms as they have the same power.
-4n^2 + (3n)^2
-4n^2 + 3^2 n^2
-4n^2 + 9n^2
5n^2

3. n^3 + 2n^2 (n + 1)

Distribute 2n^2 to n + 1. It means you have to multiply 2n^2 to the terms n and +1. Then add like terms.
n^3 + 2n^2 (n + 1)
n^3 + 2n^3 + 2n^2
3n^3 + 2n^2

4. x^3 + x^2 - 9x = 9

It's a trinomial. Transpose 9 to the other side. I have here the easy way to factor this one.
x^3 + x^2 - 9x - 9 = 0
x^2 (x + 1) - 9 (x + 1) = 0 -------> see? they have x + 1, great!
(x^2 - 9)(x + 1) = 0
Then factor x^2 - 9. Remember x^2 - y^2 = (x + y)(x - y)? There!
(x + 3)(x - 3)(x + 1)= 0
In order for this equation to be zero, set EACH of the factors equal to zero
x = -3
x = 3
x = -1

Welcome! :D

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1) Divide each side by 5:

(5n)/5 = 25/5
(5/5)n = 5
n = 5

2) First expand the (3n)^2 term:

-4n^2 + (3n)(3n)
-4n^2 + (3x3)(nxn)
-4n^2 + 9n^2

Now factor out the common n^2:

(-4 + 9)n^2
5n^2

3) First expand the 2n^2(n + 1) term then simplify:

n^3 - (2n^2 x n + 2n^2 x 1)
n^3 - (2n^3 + 2n^2)
n^3 - 2n^3 - 2n^2
-n^3 - 2n^2
-(n^3 + 2n^2)

4) This one will be much harder than the last 3 questions. What you are going to do is called "finding the zeros" (also known as "finding the roots") of this equation.

First subtract 9 from each side so that you have the right hand side equal to zero: